assume that x and y are both differentiable functions of t. find dy/dt when x=5, and dx/dt=6 given the equation x(x)+y=3

We are given the equation \( x(x) + y = 3 \) and need to find \( \frac{dy}{dt} \) when \( x = 5 \) and \( \frac{dx}{dt} = 6 \).

### Step 1: Rewrite the equation.
The equation can be simplified as:
\[
x^2 + y = 3
\]

### Step 2: Implicit differentiation with respect to \( t \).
Since both \( x \) and \( y \) are functions of \( t \), we differentiate both sides of the equation with respect to \( t \).

Differentiating \( x^2 + y = 3 \):
\[
\frac{d}{dt}(x^2) + \frac{d}{dt}(y) = \frac{d}{dt}(3)
\]
This gives:
\[
2x \cdot \frac{dx}{dt} + \frac{dy}{dt} = 0
\]

### Step 3: Solve for \( \frac{dy}{dt} \).
Rearrange the equation to isolate \( \frac{dy}{dt} \):
\[
\frac{dy}{dt} = -2x \cdot \frac{dx}{dt}
\]

### Step 4: Substitute the given values.
We are given \( x = 5 \) and \( \frac{dx}{dt} = 6 \). Plug these values into the equation:
\[
\frac{dy}{dt} = -2(5) \cdot 6 = -60
\]

Thus, \( \frac{dy}{dt} = -60 \).

Would you like more details or have any questions? 

Here are five related questions:

1. How does implicit differentiation work for more complex equations?
2. What is the role of the chain rule in implicit differentiation?
3. How do you apply implicit differentiation when there are multiple variables?
4. Can you explain why \( \frac{d}{dt}(3) = 0 \)?
5. How would this change if \( y \) were a more complex function of \( t \)?

**Tip**: Always check if the variables in an equation are functions of another variable when differentiating, as this might require implicit differentiation.

Assume that x and y are both differentiable functions of t. Find dy/dt when x=5, and dx/dt=6, given the equation x(x) + y = 3.

Solve the implicit differentiation problem where x and y are functions of t. Given the equation x^2 + y = 3, learn how to find dy/dt when x=5 and dx/dt=6. This solution includes the use of the chain rule in calculus and is suitable for students in Grades 10-12.

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